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%%文档的题目、作者与日期
\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{实变函数测验 3.1-3.2 解答}
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\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{2024 年 4 月 22 日}
%\date{March 9, 2021}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item  %Problem 01 
证明直线上的可数点集的外测度为零。
%\hfill\underline{\makebox[4cm]{理由： }}

%证明：
\begin{enumerate}[label={\arabic*.}]

\item 设可数点集为 $E=\{x_1,x_2,\cdots,x_n,\cdots\}$.   
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}可数集的定义}}

\item  给定任意正数 $\varepsilon>0$.    
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}若无限接近就只好相等 }}

\item  取开区间 $(x_1-\frac{\varepsilon}{4}, x_1+\frac{\varepsilon}{4})$ 盖住点 $x_1$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}点在区间中 }}


\item  对每个点 $x_n$, 取开区间 $(x_n-\frac{\varepsilon}{2^{n+1}}, x_n+\frac{\varepsilon}{2^{n+1}})$ 盖住这个点。 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}区间虽小，足够盖住一个点 }}

\item  点集 $E$ 的这个开覆盖的总长度为 $\frac{\varepsilon}{2}+\frac{\varepsilon}{4}+\cdots+\frac{\varepsilon}{2^n}+\cdots = \varepsilon$. 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}几何级数求和 }}

\item  因此 $m^*(E)\le \varepsilon$. 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}外测度是开覆盖的总测度的下确界 }}

\item  因此 $m^*(E)=0$. 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}6式中的 $\varepsilon$ 可以是任意正数 }}

\end{enumerate}

\vspace{0.2cm}

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\item  %Problem 02
证明外测度具有次可数可加性：对平面上的可数个点集 $A_1,A_2,\cdots, A_n,\cdots$, 有 
$$m^*(\cup_{n=1}^{\infty} A_n) \le \sum_{n=1}^{\infty}m^*(A_n). $$ 
%\hfill\underline{\makebox[4cm]{理由： }}

%证明：
\begin{enumerate}[label={\arabic*.}]

\item  给定任意正数 $\varepsilon>0$.    
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}后面证不等式用到 }}

\item  点集 $A_n$ 存在开覆盖 $\{I_{n,k}\}_{k=1}^{\infty}$ 使得 $m^*(A_i)\ge \sum\limits_{k=1}^\infty m(I_{n,k}) - \frac{\varepsilon}{2^n}$.
%\hfill \underline{\hspace{4cm}} 
\hfill\underline{\makebox[4cm]{\color{red}外测度是开覆盖的总测度的下确界 }}

\item   点集 $\cup_{n=1}^{\infty} A_n$ 有开覆盖 $\{I_{n,k}\}_{n,k=1}^{\infty}$, 所以 $m^*(\cup_{n=1}^{\infty} A_n)\le \sum\limits_{n,k=1}^\infty m(I_{n,k})$. 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}很多覆盖放一起 }}

\item   所以 $m^*(\cup_{n=1}^{\infty} A_n)\le \sum\limits_{n}^\infty \left[m^*(A_i) + \frac{\varepsilon}{2^n}\right]$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}结合上面的2与3 }}

\item   所以 $m^*(\cup_{n=1}^{\infty} A_n)\le \sum\limits_{n}^\infty m^*(A_i) + \varepsilon $.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}几何级数求和 }}

\item   所以 $m^*(\cup_{n=1}^{\infty} A_n)\le \sum\limits_{n}^\infty m^*(A_i) $.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}5式中的 $\varepsilon$ 可以是任意正数 }}


\end{enumerate}

\vspace{0.2cm}

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\item  %Problem 03
设 $E\subseteq\mathbb{R}$ 是有界点集。%设 $m^*(E)>0$. 
设 $0<c<m^*(E)$. 证明存在子集 $E_1\subseteq E$ 使得 $m^*(E_1)=c$. 
%\hfill\underline{\makebox[4cm]{理由： }}

%证明：
\begin{enumerate}[label={\arabic*.}]

\item  存在实数 $a,b\in\mathbb{R}$, 使得 $E\subseteq [a,b]$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}有界点集的定义 }}

\item  对任意 $x\in [a,b]$, 定义函数 $f(x)=m^*(E\cap [a,x])$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}外测度对所有点集都有定义 }}

\item  对任意 $x_1<x_2$, 有 $|f(x_2)-f(x_1)| = m^*(E\cap [a,x_2]) - m^*(E\cap [a,x_1])$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red} $f(x)$ 的定义、外测度是非负的 }}

\item  所以 $|f(x_2)-f(x_1)| = m^*(E\cap (x_1,x_2])$. 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}差集的运算 }}

\item  所以 $|f(x_2)-f(x_1)| \le m^*([x_1,x_2]) = x_2 - x_1$. 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}外测度是单调递增的 }}

\item  所以 $f(x)$ 是连续函数。 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}验证连续函数的定义，取 $\delta=\varepsilon$ }}

\item  $f(a)=0$, $f(b)=m^*(E)$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}代入函数 $f(x)$ 的定义 }}

\item  对 $0<c<m^*(E)$, 存在 $x_0\in [a,b]$, 使得 $f(x_0)=c$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}连续函数的介值定理 }}

\item  取 $E_1=E\cap [a,x_0]$, 则 $m^*(E_1)=c$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}函数 $f(x)$ 的定义 }}

\end{enumerate}

\vspace{0.2cm}


\newpage
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\item  %Problem 04
设 $E=[a,b]$ 为直线上的闭区间。证明外测度 $m^*(E)=b-a$.

%证明：
\begin{enumerate}[label={\arabic*.}]

\item  对任意 $\varepsilon>0$, 取开区间 $(a-\frac{\varepsilon}{2}, b+\frac{\varepsilon}{2})$ 盖住 $E$, 所以 $m^*(E)\le b-a+\varepsilon$.   
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}外测度的定义 }}

\item  所以 $m^*(E)\le b-a$.   
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}1中的 $\varepsilon$ 可以是任意正数 }}

\item  对 $E$ 的任意开覆盖 $\{(\alpha_n,\beta_n)\}_{n=1}^{\infty}$, 存在有限子覆盖 $\{(\alpha_{n_k},\beta_{n_k})\}_{k=1}^{m}$.   
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}闭区间的有限覆盖定理 }}

\item  可见 $b-a\le \sum\limits_{k=1}^{m} (\beta_{n_k} - \alpha_{n_k})$. 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}有限覆盖区间的总长度不会更短 }}

\item  所以 $m^*(E)\ge b-a$. 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}外测度的开覆盖的总测度的下确界 }}

\item  所以 $m^*(E)= b-a$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}结合2与5得到 }}


\end{enumerate}

\vspace{0.2cm}


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\item  %Problem 05
%引理。
设 $E\subseteq \mathbb{R}^2$ 是任意一个子集。证明若 $m^*(T)=m^*(T\cap E)+m^*(T\cap E^c)$ 对任意开区间 $T=I\subseteq \mathbb{R}^2$ 都成立，则这个等式对任意点集 $T\subseteq \mathbb{R}^2$ 也成立。 

%证明：
\begin{enumerate}[label={\arabic*.}]

\item  设任意点集 $T$, 设任意正数 $\varepsilon>0$.    
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red} $T$是题目要求、$\varepsilon$ 后面用到 }}

\item  存在 $T$ 的开覆盖 $\{I_n\}_{n=1}^{\infty}$ 使得 $m^*(T)\ge \sum\limits_{n=1}^{\infty} m(I_n)-\varepsilon$.   
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}外测度的开覆盖的总测度的下确界 }}

\item  从 $\{I_n\cap E\}_{n=1}^{\infty}$ 覆盖 $T\cap E$ 可得 $m^*(T\cap E)\le \sum\limits_{n=1}^{\infty} m^*(I_n\cap E)$.   
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}外测度是单调递增的、次可数可加的 }}

\item  从 $\{I_n\cap E^c\}_{n=1}^{\infty}$ 覆盖 $T\cap E^c$ 可得 $m^*(T\cap E^c)\le \sum\limits_{n=1}^{\infty} m^*(I_n\cap E^c)$.   
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}外测度是单调递增的、次可数可加的 }}

\item  所以 $m^*(T\cap E) + m^*(T\cap E^c)\le \sum\limits_{n=1}^{\infty} \left[ m^*(I_n\cap E) + m^*(I_n\cap E^c) \right] $.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}由3和4相加得到 }}

\item  所以 $m^*(T\cap E) + m^*(T\cap E^c)\le \sum\limits_{n=1}^{\infty}  m^*(I_n) $.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}题目所给条件、等式对开区间成立 }}

\item  所以 $m^*(T\cap E) + m^*(T\cap E^c)\le m^*(T) + \varepsilon$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}结合2与6得到 }}

\item  所以 $m^*(T\cap E) + m^*(T\cap E^c)\le m^*(T)$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}$\varepsilon$ 可以是任意正数 }}

\item  另一方面，又有 $m^*(T) \le m^*(T\cap E) + m^*(T\cap E^c)$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}外测度的次可加性 }}

\item  所以 $m^*(T) = m^*(T\cap E) + m^*(T\cap E^c)$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}结合8与9得到 }}


\end{enumerate}

\vspace{0.2cm}

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\item  %Problem 06
证明：若 $A,B$ 都是可测集，则 $A\cup B$ 也是可测集。

%证明：
\begin{enumerate}[label={\arabic*.}]

\item 设任意点集 $T$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}为了验证卡拉泰奥多里条件 }}

\item 因为 $A$ 可测，所以 $m^*(T) = m^*(T\cap A) + m^*(T\cap A^c)$.   
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}可测集的定义 }}

\item 因为 $B$ 可测，所以 $m^*(T\cap A^c) = m^*[(T\cap A^c)\cap B] + m^*[(T\cap A^c)\cap B^c]$.   
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}可测集的定义 }}

\item 所以 $m^*(T) = m^*(T\cap A) + m^*[(T\cap A^c)\cap B] + m^*[(T\cap A^c)\cap B^c]$.    
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}3代入2得到 }}

\item 因为 $A$ 可测，所以 
\begin{eqnarray*}
m^*[T\cap (A\cup B)] 
&=& m^*[T\cap (A\cup B)\cap A] + m^*[T\cap (A\cup B)\cap A^c] \\ 
&=& m^*(T\cap A) + m^*[(T\cap A^c)\cap B]. 
\end{eqnarray*}
%\raggedleft \hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}可测集的定义、集合的运算 }}

\item 所以  $m^*(T) = m^*[T\cap (A\cup B)] + m^*[T\cap (A\cup B)^c]$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}5代入4得到、集合的运算 }}

\item 所以  $A\cup B$ 是可测集。  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}验证了可测集的定义 }}

\end{enumerate}

\vspace{0.2cm}



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\item  %Problem 07
%H5
设 $E\subseteq\mathbb{R}^n$, 若 $m^*(E)=0$, 证明 $E$ 是可测集。

%证明：
\begin{enumerate}[label={\arabic*.}]

\item 设任意点集 $T\subseteq\mathbb{R}^n$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}为了验证卡拉泰奥多里条件 }}

\item 可得 $m^*(T)  \le  m^*(T\cap E) + m^*(T\cap E^c)  \le m^*(E) + m^*(T) = m^*(T)$. 
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}次可加性、单调递增、题目条件 }}

\item 所以 $m^*(T) = m^*(T\cap E) + m^*(T\cap E^c)$.  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}2中的小于等于只能取等于 }}

\item 所以 $E$ 是可测集。  
%\hfill \underline{\hspace{4cm}}
\hfill\underline{\makebox[4cm]{\color{red}验证了可测集的定义 }}


\end{enumerate}

\vspace{0.2cm}



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\end{enumerate}


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\end{document}

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